The gaseous state
The gaseous state
- Gas molecules move fast in all directions and collide with the walls.
- The many tiny pushes add up to the gas pressure.
- We model gases as ideal to make calculations simple.
Practice
Gas pressure is caused by:
Each collision with the wall gives a tiny push; the many pushes add up to the pressure.
Ideal vs real gases
- An ideal gas assumes: the particles take up zero volume, and there are no forces between them.
- A real gas follows this closely at low pressure and high temperature.
- It behaves least ideally at high pressure and low temperature (particles crowded, forces matter).
Practice
An ideal gas is assumed to have:
The ideal-gas model assumes point particles (zero volume) with no intermolecular forces.
Practice
A real gas behaves LEAST like an ideal gas at:
At high pressure and low temperature the particles are crowded, so their size and attractions matter.
The ideal gas equation
$$pV = nRT$$
- $p$ in Pa, $V$ in $\text{m}^3$, $n$ in mol, $T$ in kelvin (K), $R = 8.31\ \text{J}/(\text{K}\cdot\text{mol})$.
- Convert first: °C → K (add 273), and $\text{cm}^3$/$\text{dm}^3$ → $\text{m}^3$.
Practice
In pV = nRT, the temperature T must be in:
T must be in kelvin; convert °C to K by adding 273 (and volumes to m³).
Finding molar mass
Since $n = m/M$:
$$pV = \frac{m}{M}RT \quad\Rightarrow\quad M = \frac{mRT}{pV}$$
- This finds $M_r$ from the mass (or density) of a gas.
Practice
Which equation gives the molar mass of a gas?
From pV = (m/M)RT, rearranging gives M = mRT/(pV).
You've got it
Key idea
- gas pressure = sum of molecule–wall collisions
- ideal gas: zero particle volume, no forces; real gases deviate at high P, low T
- $pV = nRT$ (T in K, convert units first; $R = 8.31\ \text{J}/(\text{K}\cdot\text{mol})$)
- molar mass: $M = \dfrac{mRT}{pV}$